MHT CET · Maths · Differentiation
Let \(f\) be twice differentiable function such that \(\mathrm{f}^{\prime \prime}(x)=-\mathrm{f}(x), \mathrm{f}^{\prime}(x)=\mathrm{g}(x)\) and \(\mathrm{h}(x)=(\mathrm{f}(x))^2+(\mathrm{g}(x))^2\). If \(\mathrm{h}(5)=1\), then the value of \(h(10)\) is
- A 2
- B 1
- C \(\frac{1}{2}\)
- D -1
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
\(\mathrm{h}(x) =[\mathrm{f}(x)]^2+[\mathrm{g}(x)]^2 \)
\( \therefore \quad \mathrm{~h}^{\prime}(x) \left.=2 \mathrm{f}(x)] \mathrm{f}^{\prime}(x)+2 \mathrm{~g}(x)\right] \mathrm{g}^{\prime}(x) \)
\( =2\left[-\mathrm{f}^{\prime \prime}(x)\right] \mathrm{f}^{\prime}(x)+2\left[\mathrm{f}^{\prime}(x)\right] \cdot \mathrm{f}^{\prime \prime}(x) \)
\( =0\)
\(\therefore \mathrm{h}(x)\) is a constant function.
\(\therefore h(5)=1 \Rightarrow h(10)=1\)
\( \therefore \quad \mathrm{~h}^{\prime}(x) \left.=2 \mathrm{f}(x)] \mathrm{f}^{\prime}(x)+2 \mathrm{~g}(x)\right] \mathrm{g}^{\prime}(x) \)
\( =2\left[-\mathrm{f}^{\prime \prime}(x)\right] \mathrm{f}^{\prime}(x)+2\left[\mathrm{f}^{\prime}(x)\right] \cdot \mathrm{f}^{\prime \prime}(x) \)
\( =0\)
\(\therefore \mathrm{h}(x)\) is a constant function.
\(\therefore h(5)=1 \Rightarrow h(10)=1\)
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